Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 158191 by ghakhan88 last updated on 31/Oct/21

Can we reach to (m/((m−1)s)) + ((m+1)/(ms^2 )) from ((ms+m(m+1))/(s(m−s))) ??

$${Can}\:{we}\:{reach}\:{to}\:\frac{{m}}{\left({m}−\mathrm{1}\right){s}}\:+\:\frac{{m}+\mathrm{1}}{{ms}^{\mathrm{2}} }\:{from}\: \\ $$$$ \\ $$$$\:\:\:\:\frac{{ms}+{m}\left({m}+\mathrm{1}\right)}{{s}\left({m}−{s}\right)}\:?? \\ $$

Commented by Rasheed.Sindhi last updated on 01/Nov/21

They′re not equal even! for m=2 & s=1 they have different values. (m/((m−1)s)) + ((m+1)/(ms^2 )) =^(?) ((ms+m(m+1))/(s(m−s))) (2/((2−1)(1))) + ((2+1)/(2(1^2 ))) =^(?) ((2(1)+2(2+1))/(1(2−1))) (2/1) + (3/2) =^(?) ((2+2(3))/(2−1)) (7/2) =^(?) ((2+6)/1) (7/2)≠8

$$\mathcal{T}{hey}'{re}\:{not}\:{equal}\:{even}! \\ $$$${for}\:{m}=\mathrm{2}\:\&\:{s}=\mathrm{1}\:{they}\:{have}\:{different} \\ $$$${values}. \\ $$$$\:\frac{{m}}{\left({m}−\mathrm{1}\right){s}}\:+\:\frac{{m}+\mathrm{1}}{{ms}^{\mathrm{2}} }\:\overset{?} {=}\frac{{ms}+{m}\left({m}+\mathrm{1}\right)}{{s}\left({m}−{s}\right)} \\ $$$$\:\frac{\mathrm{2}}{\left(\mathrm{2}−\mathrm{1}\right)\left(\mathrm{1}\right)}\:+\:\frac{\mathrm{2}+\mathrm{1}}{\mathrm{2}\left(\mathrm{1}^{\mathrm{2}} \right)}\:\overset{?} {=}\frac{\mathrm{2}\left(\mathrm{1}\right)+\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right)}{\mathrm{1}\left(\mathrm{2}−\mathrm{1}\right)} \\ $$$$\:\frac{\mathrm{2}}{\mathrm{1}}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\overset{?} {\:=}\frac{\mathrm{2}+\mathrm{2}\left(\mathrm{3}\right)}{\mathrm{2}−\mathrm{1}} \\ $$$$\:\:\frac{\mathrm{7}}{\mathrm{2}}\:\overset{?} {=}\frac{\mathrm{2}+\mathrm{6}}{\mathrm{1}} \\ $$$$\frac{\mathrm{7}}{\mathrm{2}}\neq\mathrm{8} \\ $$

Answered by MJS_new last updated on 31/Oct/21

no

$$\mathrm{no} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com